Playing with Numbers — Check Your Progress

In a 2-digit number, sum of digits is 7. If the difference of 2 digit number and number obtained by reversing the digits is 9, then find the number.

Answer

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Given, sum of digits = 7

⇒ a + b = 7 .....(1)

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the difference of the 2-digit number and the number obtained by reversing the digits is 9.

⇒ (10a + b) - (10b + a) = 9

⇒ 9a - 9b = 9

⇒ 9(a - b) = 9

⇒ a - b = 1 .....(2)

Adding (1) and (2), we get

2a = 8

⇒ a = 4

Putting a = 4 in (1), we get b = 7 - 4 = 3

∴ 2-digit number = 10 × 4 + 3 = 43

If the number obtained by reversing the digits is greater than the original number, then

b - a = 1, which on solving with (1) gives a = 3 and b = 4.

∴ 2-digit number = 10 × 3 + 4 = 34

Hence, the required number is 43 or 34.

In a 3 digit number, the difference of hundred's digit and unit's digit is 5. Find the quotient when difference of 3-digit number and number obtained by reversing the digits is divided by 9.

Answer

Let the 3-digit number be abc i.e. 100a + 10b + c, where hundred's digit is a, ten's digit is b and unit's digit is c.

Given, a - c = 5

Number obtained by reversing the digits = cba i.e. 100c + 10b + a

Difference = (100a + 10b + c) - (100c + 10b + a)

= 100a + 10b + c - 100c - 10b - a

= 99a - 99c

= 99(a - c)

On dividing by 9, we get

Quotient = $\dfrac{99(a - c)}{9}$ = 11(a - c) = 11 × 5 = 55

Hence, the required quotient is 55.

Without actual calculation, write the quotient when sum of 3 digit numbers 567, 675 and 756 is divided by

(i) 111

(ii) 18

(iii) 37

(iv) 3

Answer

The numbers 675 and 756 are obtained by changing the order of digits of 567 cyclically.

According to the property, the sum of a 3-digit number abc and the numbers obtained by changing the order of digits cyclically = 111(a + b + c) = 3 × 37 × (a + b + c).

Here, a + b + c = 5 + 6 + 7 = 18

∴ 567 + 675 + 756 = 111 × 18

(i) When the sum is divided by 111, the quotient = 5 + 6 + 7 = 18.

(ii) When the sum is divided by 18 (5 + 6 + 7 = 18), the quotient = 111.

(iii) Since 111 × 18 = 37 × 3 × 18 = 37 × 54, when the sum is divided by 37, the quotient = 3 × 18 = 54.

(iv) Since 111 × 18 = 3 × 37 × 18 = 3 × 666, when the sum is divided by 3, the quotient = 37 × 18 = 666.

Find the values of the letters in the following and give reasons for the steps involved:

$\begin{matrix} & \text{A} & \text{B} & 2 \\ + & & 2 & \text{A} \\ \hline & \text{A} & \text{A} & 8 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & \text{A} & \text{B} & 2 \\ + & & 2 & \text{A} \\ \hline & \text{A} & \text{A} & 8 \\ \hline \end{matrix}$

Firstly, we will find the value of A.

Clearly, 2 + A is a number whose ones digit is 8.

⇒ 2 + A = 8, 2 + A = 18; and so on.

⇒ A = 8 - 2, A = 18 - 2; and so on.

⇒ A = 6, A = 16; and so on.

Since, A is a digit. ∴ A = 6 and there is no carry over.

Secondly, we will find the value of B.

Ten's column: B + 2 is a number whose ones digit is A i.e. 6.

⇒ B + 2 = 6 or B + 2 = 16

⇒ B = 4 or B = 14

Since, B is a digit. ∴ B = 4 and there is no carry over.

Hundred's column: A = A, which is correct.

$\begin{matrix} & \text{6} & \text{4} & 2 \\ + & & 2 & \text{6} \\ \hline & \text{6} & \text{6} & 8 \\ \hline \end{matrix}$

Hence, A = 6 and B = 4.

Find the values of the letters in the following and give reasons for the steps involved:

$\begin{matrix} & & 4 & \text{A} & \text{B} \\ + & & \text{A} & \text{B} & 7 \\ \hline & 1 & 0 & 2 & 3 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & & 4 & \text{A} & \text{B} \\ + & & \text{A} & \text{B} & 7 \\ \hline & 1 & 0 & 2 & 3 \\ \hline \end{matrix}$

Firstly, we will find the value of B.

Clearly, B + 7 is a number whose ones digit is 3.

⇒ B + 7 = 3, B + 7 = 13, B + 7 = 23; and so on.

⇒ B = 3 - 7, B = 13 - 7, B = 23 - 7; and so on.

⇒ B = -4, B = 6, B = 16; and so on.

Since, B is a digit. ∴ B = 6 and 1 is carried over to ten's column.

Secondly, we will find the value of A.

Ten's column: A + B + 1 = A + 6 + 1 = A + 7 is a number whose ones digit is 2.

⇒ A + 7 = 2, A + 7 = 12, A + 7 = 22; and so on.

⇒ A = 2 - 7, A = 12 - 7, A = 22 - 7; and so on.

⇒ A = -5, A = 5, A = 15; and so on.

Since, A is a digit. ∴ A = 5 and 1 is carried over to hundred's column.

Hundred's column: 4 + A + 1 = 4 + 5 + 1 = 10, write 0 and 1 is carried over to thousand's column, which is correct.

$\begin{matrix} & & \overset{1}{4} & \overset{1}{\text{5}} & \text{6} \\ + & & \text{5} & \text{6} & 7 \\ \hline & 1 & 0 & 2 & 3 \\ \hline \end{matrix}$

Hence, A = 5 and B = 6.

Find the values of the letters in the following and give reasons for the steps involved:

$\begin{matrix} & \text{A} & \text{A} & \text{B} \\ \times & & & \text{B} \\ \hline & 8 & 8 & \text{A} \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & \text{A} & \text{A} & \text{B} \\ \times & & & \text{B} \\ \hline & 8 & 8 & \text{A} \\ \hline \end{matrix}$

Clearly, B × B is a number whose ones digit is A.

Also, AAB × B = 88A, a 3-digit number lying between 880 and 889.

Let's take B = 1: Then A = 1 (as 1 × 1 = 1) and 111 × 1 = 111, which is not of the form 88A. Hence, B ≠ 1.

Let's take B = 2: Then A = 4 (as 2 × 2 = 4) and 442 × 2 = 884, which is of the form 88A with A = 4.

Let's take B = 3: Then A = 9 (as 3 × 3 = 9) and 993 × 3 = 2979, which is a 4-digit number. Hence, B ≠ 3.

Similarly, for B = 4, 5, 6, 7, 8 or 9, the product AAB × B becomes a 4-digit number. Hence, these values are not possible.

$\begin{matrix} & \text{4} & \text{4} & \text{2} \\ \times & & & \text{2} \\ \hline & 8 & 8 & \text{4} \\ \hline \end{matrix}$

Hence, A = 4 and B = 2.

If $923x783$ is divisible by 11, what is the value of digit $x$?

Answer

According to the divisibility of 11, if the difference of the sum of digits at the odd places (starting from unit's place) and the sum of the digits at even places is either 0 or divisible by 11, then the number is divisible by 11.

For the number 923x783,

Sum of digits at odd places = 3 + 7 + 3 + 9 = 22

Sum of digits at even places = 8 + x + 2 = 10 + x

Their difference = 22 - (10 + x) = 12 - x, which should be either 0 or divisible by 11.

⇒ 12 - x = 0, 11, 22, ...............

⇒ x = 12, 1, -10, ...............

Since x is a digit, therefore x = 1.

Hence, the value of digit x is 1.

Check the divisibility of the following number by 2, 3, 9 and 11:

76543

Answer

Last digit is 3, which is an odd number, so it is not divisible by 2.

Sum of digits = 7 + 6 + 5 + 4 + 3 = 25, which is neither divisible by 3 nor by 9, so it is not divisible by 3 or 9.

Sum of digits at odd places = 3 + 5 + 7 = 15 and sum of digits at even places = 4 + 6 = 10. Their difference = 15 - 10 = 5, which is neither 0 nor divisible by 11, so it is not divisible by 11.

Hence, 76543 is not divisible by 2, 3, 9 or 11.

Check the divisibility of the following number by 2, 3, 9 and 11:

65432

Answer

Last digit is 2, so it is divisible by 2.

Sum of digits = 6 + 5 + 4 + 3 + 2 = 20, which is neither divisible by 3 nor by 9, so it is not divisible by 3 or 9.

Sum of digits at odd places = 2 + 4 + 6 = 12 and sum of digits at even places = 3 + 5 = 8. Their difference = 12 - 8 = 4, which is neither 0 nor divisible by 11, so it is not divisible by 11.

Hence, 65432 is divisible by 2 only.

Check the divisibility of the following number by 2, 3, 9 and 11:

98765436

Answer

Last digit is 6, so it is divisible by 2.

Sum of digits = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 6 = 48, which is divisible by 3 but not by 9, so it is divisible by 3 but not by 9.

Sum of digits at odd places = 6 + 4 + 6 + 8 = 24 and sum of digits at even places = 3 + 5 + 7 + 9 = 24. Their difference = 24 - 24 = 0, so it is divisible by 11.

Hence, 98765436 is divisible by 2, 3 and 11 but not by 9.

Check the divisibility of the following number by 2, 3, 9 and 11:

234567

Answer

Last digit is 7, which is an odd number, so it is not divisible by 2.

Sum of digits = 2 + 3 + 4 + 5 + 6 + 7 = 27, which is divisible by both 3 and 9, so it is divisible by 3 and 9.

Sum of digits at odd places = 7 + 5 + 3 = 15 and sum of digits at even places = 6 + 4 + 2 = 12. Their difference = 15 - 12 = 3, which is neither 0 nor divisible by 11, so it is not divisible by 11.

Hence, 234567 is divisible by 3 and 9 but not by 2 or 11.

Check the divisibility of the following number by 5 or 10:

23565

Answer

According to the divisibility of 5, if the unit's digit of the number is 0 or 5 then the number is divisible by 5.

According to the divisibility of 10, if the unit's digit of the number is 0 then the number is divisible by 10.

Last digit of 23565 is 5. So, it is divisible by 5 but not by 10.

Hence, 23565 is divisible by 5 but not by 10.

Check the divisibility of the following number by 5 or 10:

45270

Answer

According to the divisibility of 5, if the unit's digit of the number is 0 or 5 then the number is divisible by 5.

According to the divisibility of 10, if the unit's digit of the number is 0 then the number is divisible by 10.

Last digit of 45270 is 0. So, it is divisible by both 5 and 10.

Hence, 45270 is divisible by both 5 and 10.

Check the divisibility of the following number by 4 or 8:

47596

Answer

According to the divisibility of 4, if the number formed by its last two (ten's and unit's) digits is divisible by 4 then the number is divisible by 4.

According to the divisibility of 8, if the number formed by its last three (hundred's, ten's and unit's) digits is divisible by 8 then the number is divisible by 8.

Number formed by last two digits of 47596 = 96 = 4 × 24, so it is divisible by 4. Number formed by last three digits = 596, which is not divisible by 8. So, it is not divisible by 8.

Hence, 47596 is divisible by 4 but not by 8.

Check the divisibility of the following number by 4 or 8:

593024

Answer

According to the divisibility of 4, if the number formed by its last two (ten's and unit's) digits is divisible by 4 then the number is divisible by 4.

According to the divisibility of 8, if the number formed by its last three (hundred's, ten's and unit's) digits is divisible by 8 then the number is divisible by 8.

Number formed by last two digits of 593024 = 24 = 4 × 6, so it is divisible by 4. Number formed by last three digits = 024 = 8 × 3, so it is divisible by 8.

Hence, 593024 is divisible by both 4 and 8.

If the difference of two digit number and number obtained by reversing the digits is 45, then write all possible 2-digit numbers.

Answer

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the difference of the 2-digit number and the number obtained by reversing the digits is 45.

⇒ (10a + b) - (10b + a) = 45

⇒ 9a - 9b = 45

⇒ 9(a - b) = 45

⇒ a - b = 5

The possible pairs of digits with a - b = 5 are:

a = 6, b = 1; a = 7, b = 2; a = 8, b = 3; a = 9, b = 4

(When a = 5 and b = 0, the number obtained by reversing the digits of 50 i.e. 05 is not a 2-digit number, so it is not considered.)

∴ The possible numbers are 61, 72, 83, 94 and the numbers obtained by reversing their digits are 16, 27, 38, 49.

Hence, all possible 2-digit numbers are 16, 61; 27, 72; 38, 83; 49, 94.